Question

A block $A$ of mass $m_1$ rests on a horizontal table. $A$ light string connected to it passes over a frictionless pully at the edge of table and from its other end another block $B$ of mass $m_2$ is suspended. The coefficient of kinetic friction between the block and the table is ${\mu }_k$. When the block $A$ is sliding on the table, the tension in the string is

• A. $\frac{m_1{m}_2(1+{\mu }_k)g}{(m_1+m_2)}$
• B. $\frac{m_1{m}_2(1-{\mu }_k)g}{(m_1+m_2)}$
• C. $\frac{(m_1+{\mu }_k{m}_1)g}{(m_1+m_2)}$
• D. $\frac{(m_2-{\mu }_k{m}_1)g}{(m_1+m_2)}$

Answer 1

Answer: (A)

The blocks $m_1$ and $m_2$ will move with combined acceleration a:
From F.B.D. of block $m_1$
$T-f_1=m_{1}a\dots (i)$
as the block $m_1$ is sliding, kinetic friction will be acting:
$T-{\mu }_{k}N=m_{1}a\dots$ (ii)
$N=mg\dots$ (iii)
From F.B.D. of block $m_2$
$m_{2}g-T=m_{2}a\dots$ (iv)
adding (ii) and (iv)
$a=\frac{m_{2}g-{\mu }_{k}N}{m_1+m_2}\dots (v)$
$T=m_{2}g-m_{2}a=m_2\left(g-\frac{m_{2}g-{\mu }_{k}N}{m_1+m_2}\right)$
$=m_2\left(\frac{m_{1}g+{\mu }_k{m}_{1}g}{m_1+m_2}\right)$
$=m_1{m}_2\left(\frac{1+{\mu }_k}{m_1+m_2}g\right)$